# Multiplicity of an eigenvalue is equal to $\dim V_{\lambda}$

I am trying to prove that multiplicity of an eigenvaliue $\lambda$ = $\dim V_{\lambda}$ and I have problems with this inequality:

$\dim V_{\lambda} \le$ multiplicity $\lambda$.

I know that characyeristic polynomial of $W_f(t)=f|_{V_{\lambda}}$ (where $f$ has Jordan form) is $\pm(t-\lambda)^{dimV_{\lambda}}$ because there are only $\lambda$'s on the matrix's diagonal.

I also need to prove that $\pm(t-\lambda)^{dimV_{\lambda}}$ divides $W_f(t)$, but I don't know how. Could you help me with that?

$V_{\lambda} = \bigcup_{k=0} ^{\infty} V_{\lambda} ^{k}$, $V_{\lambda} ^{k} = ker((f-\lambda)^k)$

By multiplicity I mean algebraic multiplicity.

• What exactly do you mean by multiplicity of an eigenvalue, and what by $V_\lambda$? For example the multiplicity could refer to the power in the characteristic or the minimal polynomial. $V_\lambda$ could be the eigenspace or the generalised eigenspace. – Simon Markett May 22 '13 at 9:08
• I've already corrected my question. – Sandy May 22 '13 at 9:16
• $V_{\lambda}$ should be given by a sum, not a union. – Christopher A. Wong May 22 '13 at 9:50
• What do you mean by that? – Sandy May 22 '13 at 9:56
• @ChristopherA.Wong, the $V_\lambda^k$ are nested, so the union is fine. – Simon Markett May 22 '13 at 10:10

Hint: Over the complex numbers you have both,

$$\sum_\lambda \dim V_\lambda=n$$

and

$$\sum_\lambda mult(\lambda)=n$$

where $n$ is the size of the matrix.

• I'm sorry, but I still don't see how to use it to prove that algebraic multiplicity is equal to geometric multiplicity. – Sandy May 22 '13 at 14:49
• Didn't you say that you already showed the inequality in one direction? If you have $mult(\lambda)\leq dim V_\lambda$ for all eigenvalues, but the sum has to be equal... – Simon Markett May 22 '13 at 14:53
• Ok, I've already figured that out. Thanks. – Sandy May 22 '13 at 16:00